To find the magnitude and direction of the net electric force on each of the charged spheres, we'll use Coulomb's law. Coulomb's law gives the force between two point charges as:

$F = k_e \frac{|q_1 q_2|}{r^2}$

Where:

• $k_e$ is the Coulomb's constant, $k_e = 8.99 \times 10^9 \ \text{N m}^2/\text{C}^2$
• $q_1$ and $q_2$ are the charges of the two spheres (in Coulombs)
• $r$ is the distance between the two charges (in meters)

Given:

• $q_A = 6.10 \ \text{nC} = 6.10 \times 10^{-9} \ \text{C}$
• $q_B = 1.60 \ \text{nC} = 1.60 \times 10^{-9} \ \text{C}$
• $q_C = -2.50 \ \text{nC} = -2.50 \times 10^{-9} \ \text{C}$
• Distance between A and B: $r_{AB} = 3.00 \ \text{cm} = 0.0300 \ \text{m}$
• Distance between B and C: $r_{BC} = 2.00 \ \text{cm} = 0.0200 \ \text{m}$
• Distance between A and C: $r_{AC} = r_{AB} + r_{BC} = 0.0500 \ \text{m}$

Steps to Calculate Forces:

1. Force on B due to A ($F_{AB}$): $F_{AB} = k_e \frac{|q_A \cdot q_B|}{r_{AB}^2}$

2. Force on B due to C ($F_{BC}$): $F_{BC} = k_e \frac{|q_B \cdot q_C|}{r_{BC}^2}$

3. Force on A due to B ($F_{AB}$), same as calculated above, but with different direction.

4. Force on A due to C ($F_{AC}$): $F_{AC} = k_e \frac{|q_A \cdot q_C|}{r_{AC}^2}$

5. Force on C due to B ($F_{CB}$), same as $F_{BC}$ but direction changes.

6. Force on C due to A ($F_{CA}$), same as $F_{AC}$.

Next, calculate each force and sum them up considering their directions.

Let’s compute each force:

Step 1: Calculate the force on B due to A ($F_{AB}$):

$F_{AB} = 8.99 \times 10^9 \times \frac{(6.10 \times 10^{-9})(1.60 \times 10^{-9})}{(0.0300)^2}$

Step 2: Calculate the force on B due to C ($F_{BC}$):

$F_{BC} = 8.99 \times 10^9 \times \frac{(1.60 \times 10^{-9})(2.50 \times 10^{-9})}{(0.0200)^2}$

Step 3: Calculate force on A due to C ($F_{AC}$):

$F_{AC} = 8.99 \times 10^9 \times \frac{(6.10 \times 10^{-9})(2.50 \times 10^{-9})}{(0.0500)^2}$

After calculating these, sum the forces considering directions and compute the net force on each sphere.

Let me do the calculations.The calculated forces are:

1. Force on B due to A ($F_{AB}$): $9.75 \times 10^{-5} \ \text{N}$ to the right.
2. Force on B due to C ($F_{BC}$): $8.99 \times 10^{-5} \ \text{N}$ to the left.
3. Force on A due to C ($F_{AC}$): $5.48 \times 10^{-5} \ \text{N}$ to the left.

Net Force on Each Sphere:

1. Sphere A: The net force on A is due to B (right) and C (left).
2. Sphere B: The net force on B is the sum of the forces from A (right) and C (left).
3. Sphere C: The net force on C is due to A (right) and B (left).

Let’s summarize the net forces and directions using the vector sum of these forces.